ngu quá có thế cũng không làm được

Nguyễn Minh Phương trẻ trâu quá giỏi làm đi ko làm đc thì câm ko làm đc mà  oai thì ăn chửi

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)

\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)

\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)

\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xy+yz\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)

Mà \(x\ne y\)nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)

\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Đpcm

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)

\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)

\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)

\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)

\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)

\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)

\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)

\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)

Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)

áp dụng tính chất của dãy tỉ số bằng nhau lf ra

Áp dụng t/c dãy tỉ số bằng nhau có:

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x-xyz-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

=> \(\frac{x^2-yz}{x\left(1-yz\right)}=x+y+z\)

<=> \(\frac{x^2-yz}{x\left(1-yz\right)}-\frac{\left(x+y+z\right)x\left(1-yz\right)}{x\left(1-yz\right)}=0\)

<=> \(\frac{x^2-yz-\left(x^2+yx+zx\right)\left(1-yz\right)}{x\left(1-yz\right)}\)=0

<=> \(x^2-yz-x^2+x^2yz-xy+xy^2z-xz+xyz^2=0\)

<=> \(-yz-xy-xz+xyz\left(x+y+z\right)\)=0

<=> \(xyz\left(x+y+z\right)=yz+xy+xz\)

<=>\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)( chia cả hai vế cho xyz với x,y,z khác 0)

Ta có \(xy+xz+yz=xyz\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z=\frac{xy+xz+yz}{xyz}\left(1\right)\)

Ta lại có \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x\left(1-yz\right)-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Vậy ta có đpcm